Question

The unit vector which is orthogonal to the vector $3\hat{i}+2\hat{j}+6\hat{k}$ is coplanar with vectors $2\hat{i}+\hat{j}+6\hat{k}and\hat{i}-\hat{j}-\hat{k}$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\left(\overrightarrow{a}\right)\times (\overrightarrow{b}\times \overrightarrow{c})$ is cross product of two vectors $\left(\overrightarrow{a}\right)and(\overrightarrow{b}\times \overrightarrow{c})$.
So this vector is orthogonal to $\overrightarrow{a}$ and coplanar with $\overrightarrow{b}and\overrightarrow{c}$.
So the vector orthogonal to $3\hat{i}+2\hat{j}+6\hat{k}$ and coplanar with $2\hat{i}+\hat{j}+\hat{k}and\hat{i}-\hat{j}-\hat{k}$ will be given by $(3\hat{i}+2\hat{j}+6\hat{k})\times \left(\right(2\hat{i}+\hat{j}+\hat{k})\times (\hat{i}-\hat{j}-\hat{k}\left)\right)$
Vector triple product
Let $3\hat{i}+2\hat{j}+6\hat{k}=\overrightarrow{a}$
$2\hat{i}+\hat{j}+\hat{k}=\overrightarrow{b}$
$\hat{i}-\hat{j}-\hat{k}=\overrightarrow{c}$
So we have to calculate $\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})$
$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}.\overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{c}$
$(\overrightarrow{a}.\overrightarrow{c})=(3\hat{i}+2\hat{j}+6\hat{k}).(\hat{i}-\hat{j}-\hat{k})$
= 3 - 2 + 6
= 7
$(\overrightarrow{a}.\overrightarrow{b})=(3\hat{i}+2\hat{j}+6\hat{k}).(2\hat{i}+\hat{j}+\hat{k})$ = 6 + 2 + 6
= 14
So $\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=7(2\hat{i}+\hat{j}+\hat{k})-14(\hat{i}-\hat{j}-\hat{k})$
$=21\hat{j}-7\hat{k}$
So the unit vector for this vector will be obtained by dividing it by its magnitude
Magnitude of $21\hat{j}-7\hat{k}=\sqrt{(21{)}^2+(7{)}^2}$
$=\sqrt{490}$
$=7\sqrt{10}$
So unit vector of $21\hat{j}-7\hat{k}$
$=\frac{21\hat{j}-7\hat{k}}{7\sqrt{10}}$
$=\frac{7(3\hat{j}-7\hat{k})}{7\sqrt{10}}$
$=\frac{3\hat{j}-\hat{k}}{\sqrt{10}}.$